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3 years ago

Which expression is equivalent to

Mathematics

1 answer:

pychu [463]3 years ago

8 0

Answer:

Step-by-step explanation:

i cannot find an answer from the expression but i can guess if you didnt follow pemdas, it would be d

3.5 × 102 is 357

(8 × 10−4) + (6 × 10−4) is 132

132/357= .369 or .37 or 0 simplified

d is 4*10-10

10-10 is 0

4*0 is 0

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Is 8/10=2/5 a true proportion

FinnZ [79.3K]

No, a true proportion states that two ratios are equal.
8/10 = 4/5
2/5 = 4/10

5 0

3 years ago

Factor the polynomial and use the zeros to identify its graph. x4 − 5x3 − 20x2 + 60x + 144

Monica [59]

The zeros of the polynomial are -3, -2, 4 and 6 and the graph of the polynomial is graph (a)

<h3>How to factor the polynomial?</h3>

The polynomial is given as:

P(x) = x^4 - 5x^3 - 20x^2 + 60x + 144

Expand the polynomial function

P(x) = x^4 - 10x^3 + 5x^3 + 24x^2 - 50x^2 + 6x^2 + 120x - 60x + 144

Rewrite the function as:

P(x) = x^4 - 10x^3 + 24x^2 + 5x^3 - 50x^2 + 120x + 6x^2 - 60x + 144

Factorize the function

P(x) = x^2(x^2 - 10x + 24) + 5x(x^2 - 10x + 24) + 6(x^2 - 10x + 24)

Factor out x^2 - 10x + 24

P(x) = (x^2 + 5x + 6)(x^2 - 10x + 24)

Expand each bracket

P(x) = (x^2 + 3x + 2x + 6)(x^2 - 4x - 6x + 24)

Factorize each bracket

P(x) = [x(x + 3) + 2(x + 3)][x(x - 6) - 4(x - 6)]

Factor out x + 3 and x - 6

P(x) = (x + 3)(x + 2)(x - 4)(x - 6)

Set to 0

(x + 3)(x + 2)(x - 4)(x - 6) = 0

Solve for x

x =-3, -2, 4 and 6

Hence, the zeros of the polynomial are -3, -2, 4 and 6 and the graph of the polynomial is graph (a)

Read more about polynomials at:

brainly.com/question/20896994

5 0

2 years ago

Which whole number property is represented?

Mademuasel [1]

Associative Property of Multiplication

Because the grouping is the only things that changes

4 0

3 years ago

Root 3 cosec140° - sec140°=4<br>prove that<br><br>

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

<u>Proof:</u>

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$